HC Verma Solutions: Chapter 26 - Laws of Thermodynamics | Physics Class 11 - NEET PDF Download

 Page 1


26.1
CHAPTER 26
LAWS OF THERMODYNAMICS
QUESTIONS FOR SHORT ANSWER
1. No in isothermal process heat is added to a system. The temperature does not increase so the internal 
energy does not.
2. Yes, the internal energy must increase when temp. increases; as internal energy depends upon 
temperature U ? T
3. Work done on the gas is 0. as the P.E. of the container si increased and not of gas. Work done by the 
gas is 0. as the gas is not expanding.
The temperature of the gas is decreased.
4. W = F × d = Fd Cos 0° = Fd
Change in PE is zero. Change in KE is non Zero. 
So, there may be some internal energy.
5. The outer surface of the cylinder is rubbed vigorously by a polishing machine. 
The energy given to the cylinder is work. The heat is produced on the cylinder which transferred to the 
gas.
6. No. work done by rubbing the hands in converted to heat and the hands become warm.
7. When the bottle is shaken the liquid in it is also shaken. Thus work is done on the liquid. But heat is not 
transferred to the liquid.
8. Final volume = Initial volume. So, the process is isobaric.
Work done in an isobaric process is necessarily zero.
9. No word can be done by the system without changing its volume.
10. Internal energy = U = nC
V
T
Now, since gas is continuously pumped in. So n
2
= 2n
1
as the p
2
= 2p
1
. Hence the internal energy is 
also doubled.
11. When the tyre bursts, there is adiabatic expansion of the air because the pressure of the air inside is 
sufficiently higher than atmospheric pressure. In expansion air does some work against surroundings. 
So the internal energy decreases. This leads to a fall in temperature.
12. ‘No’, work is done on the system during this process. No, because the object expands during the 
process i.e. volume increases.
13. No, it is not a reversible process.
14. Total heat input = Total heat out put i.e., the total heat energy given to the system is converted to 
mechanical work.
15. Yes, the entropy of the body decreases. But in order to cool down a body we need another external sink 
which draws out the heat the entropy of object in partly transferred to the external sink. Thus once 
entropy is created. It is kept by universe. And it is never destroyed. This is according to the 2
nd
law of  
thermodynamics
OBJECTIVE – ?
1. (d) Dq = DU + DW. This is the statement of law of conservation of energy. The energy provided is 
utilized to do work as well as increase the molecular K.E. and P.E.
2. (b) Since it is an isothermal process. So temp. will remain constant as a result ‘U’ or internal energy will 
also remain constant. So the system has to do positive work.
3. (a) In case of A ?W
1
> ?W
2
(Area under the graph is higher for A than for B). 
?Q = ?u + dw.
du for both the processes is same (as it is a state function) 
??Q
1
> ?Q
2
  as ?W
1
> ?W
2
4. (b) As Internal energy is a state function and not a path function. ?U
1
= ?U
2
1
F
d 1
V
B ?Q 2 ?
P
A ?Q 1 ?
V
P
A
B
Page 2


26.1
CHAPTER 26
LAWS OF THERMODYNAMICS
QUESTIONS FOR SHORT ANSWER
1. No in isothermal process heat is added to a system. The temperature does not increase so the internal 
energy does not.
2. Yes, the internal energy must increase when temp. increases; as internal energy depends upon 
temperature U ? T
3. Work done on the gas is 0. as the P.E. of the container si increased and not of gas. Work done by the 
gas is 0. as the gas is not expanding.
The temperature of the gas is decreased.
4. W = F × d = Fd Cos 0° = Fd
Change in PE is zero. Change in KE is non Zero. 
So, there may be some internal energy.
5. The outer surface of the cylinder is rubbed vigorously by a polishing machine. 
The energy given to the cylinder is work. The heat is produced on the cylinder which transferred to the 
gas.
6. No. work done by rubbing the hands in converted to heat and the hands become warm.
7. When the bottle is shaken the liquid in it is also shaken. Thus work is done on the liquid. But heat is not 
transferred to the liquid.
8. Final volume = Initial volume. So, the process is isobaric.
Work done in an isobaric process is necessarily zero.
9. No word can be done by the system without changing its volume.
10. Internal energy = U = nC
V
T
Now, since gas is continuously pumped in. So n
2
= 2n
1
as the p
2
= 2p
1
. Hence the internal energy is 
also doubled.
11. When the tyre bursts, there is adiabatic expansion of the air because the pressure of the air inside is 
sufficiently higher than atmospheric pressure. In expansion air does some work against surroundings. 
So the internal energy decreases. This leads to a fall in temperature.
12. ‘No’, work is done on the system during this process. No, because the object expands during the 
process i.e. volume increases.
13. No, it is not a reversible process.
14. Total heat input = Total heat out put i.e., the total heat energy given to the system is converted to 
mechanical work.
15. Yes, the entropy of the body decreases. But in order to cool down a body we need another external sink 
which draws out the heat the entropy of object in partly transferred to the external sink. Thus once 
entropy is created. It is kept by universe. And it is never destroyed. This is according to the 2
nd
law of  
thermodynamics
OBJECTIVE – ?
1. (d) Dq = DU + DW. This is the statement of law of conservation of energy. The energy provided is 
utilized to do work as well as increase the molecular K.E. and P.E.
2. (b) Since it is an isothermal process. So temp. will remain constant as a result ‘U’ or internal energy will 
also remain constant. So the system has to do positive work.
3. (a) In case of A ?W
1
> ?W
2
(Area under the graph is higher for A than for B). 
?Q = ?u + dw.
du for both the processes is same (as it is a state function) 
??Q
1
> ?Q
2
  as ?W
1
> ?W
2
4. (b) As Internal energy is a state function and not a path function. ?U
1
= ?U
2
1
F
d 1
V
B ?Q 2 ?
P
A ?Q 1 ?
V
P
A
B
Laws of thermodynamics
26.2
5. (a) In the process the volume of the system increases continuously. Thus, the work 
done increases continuously.
6. (c) for A ? In a so thermal system temp remains same although heat is added. 
for B ? For the work done by the system volume increase as is consumes heat.
7. (c) In this case P and T varry proportionally i.e. P/T = constant. This is possible only 
when volume does not change. ? pdv = 0 ?
8. (c) Given : ?V
A
= ?V
B
. But P
A
< P
B
Now, W
A
= P
A
?V
B
; W
B
= P
B
?V
B
; So, W
A
< W
B
. 
?
9. (b) As the volume of the gas decreases, the temperature increases as well as the pressure. But, on 
passage of time, the heat develops radiates through the metallic cylinder thus T decreases as well as 
the pressure.
OBJECTIVE – ??
?
1. (b), (c) Pressure P and Volume V both increases. Thus work done is positive (V increases). Heat must 
be added to the system to follow this process. So temperature must increases.
2. (a) (b) Initial temp = Final Temp. Initial internal energy = Final internal energy.
i.e. ?U = 0, So, this is found in case of a cyclic process.
3. (d) ?U = Heat supplied, ?W = Work done.
( ?Q – ?W) = du, du is same for both the methods since it is a state function. ?
4. (a) (c) Since it is a cyclic process.
So, ?U
1
= – ?U
2
, hence ?U
1
+ ?U
2
= 0
?Q – ?W = 0 ?
5. (a) (d) Internal energy decreases by the same amount as work done. 
du = dw, ? dQ = 0. Thus the process is adiabatic. In adiabatic process, dU = – dw. Since ‘U’ decreases 
U
2
– U
2
is –ve. ?dw should be +ve ? ? ?
2 1
T T
1
nR
?
? ?
is +ve. T
1
> T
2
  ? Temperature decreases.
EXERCISES
1. t
1
= 15°c t
2
= 17°c
?t = t
2
– t
1
= 17 – 15 = 2°C = 2 + 273 = 275 K
m
v
= 100 g = 0.1 kg m
w
= 200 g = 0.2 kg
cu
g
= 420 J/kg–k W
g
= 4200 J/kg–k
(a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the 
vessel and water.
(b) Work done on the system = Heat produced unit
? dw = 100 × 10
–3
× 420 × 2 + 200 × 10
–3
× 4200 × 2 = 84 + 84 × 20 = 84 × 21 = 1764 J.
(c)dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]
2. (a) Heat is not given to the liquid. Instead the mechanical work done is converted 
to heat. So, heat given to liquid is z.
(b) Work done on the liquid is the PE lost by the 12 kg mass = mgh = 12 × 10× 
0.70 = 84 J
(c) Rise in temp at ?t We know, 84 = ms ?t 
? 84 = 1 × 4200 × ?t (for ‘m’ = 1kg) ? ?t = 
4200
84
= 0.02 k ?
V
P
f
T
P
A
B
T
P
A
B
V
P
12 kg
Page 3


26.1
CHAPTER 26
LAWS OF THERMODYNAMICS
QUESTIONS FOR SHORT ANSWER
1. No in isothermal process heat is added to a system. The temperature does not increase so the internal 
energy does not.
2. Yes, the internal energy must increase when temp. increases; as internal energy depends upon 
temperature U ? T
3. Work done on the gas is 0. as the P.E. of the container si increased and not of gas. Work done by the 
gas is 0. as the gas is not expanding.
The temperature of the gas is decreased.
4. W = F × d = Fd Cos 0° = Fd
Change in PE is zero. Change in KE is non Zero. 
So, there may be some internal energy.
5. The outer surface of the cylinder is rubbed vigorously by a polishing machine. 
The energy given to the cylinder is work. The heat is produced on the cylinder which transferred to the 
gas.
6. No. work done by rubbing the hands in converted to heat and the hands become warm.
7. When the bottle is shaken the liquid in it is also shaken. Thus work is done on the liquid. But heat is not 
transferred to the liquid.
8. Final volume = Initial volume. So, the process is isobaric.
Work done in an isobaric process is necessarily zero.
9. No word can be done by the system without changing its volume.
10. Internal energy = U = nC
V
T
Now, since gas is continuously pumped in. So n
2
= 2n
1
as the p
2
= 2p
1
. Hence the internal energy is 
also doubled.
11. When the tyre bursts, there is adiabatic expansion of the air because the pressure of the air inside is 
sufficiently higher than atmospheric pressure. In expansion air does some work against surroundings. 
So the internal energy decreases. This leads to a fall in temperature.
12. ‘No’, work is done on the system during this process. No, because the object expands during the 
process i.e. volume increases.
13. No, it is not a reversible process.
14. Total heat input = Total heat out put i.e., the total heat energy given to the system is converted to 
mechanical work.
15. Yes, the entropy of the body decreases. But in order to cool down a body we need another external sink 
which draws out the heat the entropy of object in partly transferred to the external sink. Thus once 
entropy is created. It is kept by universe. And it is never destroyed. This is according to the 2
nd
law of  
thermodynamics
OBJECTIVE – ?
1. (d) Dq = DU + DW. This is the statement of law of conservation of energy. The energy provided is 
utilized to do work as well as increase the molecular K.E. and P.E.
2. (b) Since it is an isothermal process. So temp. will remain constant as a result ‘U’ or internal energy will 
also remain constant. So the system has to do positive work.
3. (a) In case of A ?W
1
> ?W
2
(Area under the graph is higher for A than for B). 
?Q = ?u + dw.
du for both the processes is same (as it is a state function) 
??Q
1
> ?Q
2
  as ?W
1
> ?W
2
4. (b) As Internal energy is a state function and not a path function. ?U
1
= ?U
2
1
F
d 1
V
B ?Q 2 ?
P
A ?Q 1 ?
V
P
A
B
Laws of thermodynamics
26.2
5. (a) In the process the volume of the system increases continuously. Thus, the work 
done increases continuously.
6. (c) for A ? In a so thermal system temp remains same although heat is added. 
for B ? For the work done by the system volume increase as is consumes heat.
7. (c) In this case P and T varry proportionally i.e. P/T = constant. This is possible only 
when volume does not change. ? pdv = 0 ?
8. (c) Given : ?V
A
= ?V
B
. But P
A
< P
B
Now, W
A
= P
A
?V
B
; W
B
= P
B
?V
B
; So, W
A
< W
B
. 
?
9. (b) As the volume of the gas decreases, the temperature increases as well as the pressure. But, on 
passage of time, the heat develops radiates through the metallic cylinder thus T decreases as well as 
the pressure.
OBJECTIVE – ??
?
1. (b), (c) Pressure P and Volume V both increases. Thus work done is positive (V increases). Heat must 
be added to the system to follow this process. So temperature must increases.
2. (a) (b) Initial temp = Final Temp. Initial internal energy = Final internal energy.
i.e. ?U = 0, So, this is found in case of a cyclic process.
3. (d) ?U = Heat supplied, ?W = Work done.
( ?Q – ?W) = du, du is same for both the methods since it is a state function. ?
4. (a) (c) Since it is a cyclic process.
So, ?U
1
= – ?U
2
, hence ?U
1
+ ?U
2
= 0
?Q – ?W = 0 ?
5. (a) (d) Internal energy decreases by the same amount as work done. 
du = dw, ? dQ = 0. Thus the process is adiabatic. In adiabatic process, dU = – dw. Since ‘U’ decreases 
U
2
– U
2
is –ve. ?dw should be +ve ? ? ?
2 1
T T
1
nR
?
? ?
is +ve. T
1
> T
2
  ? Temperature decreases.
EXERCISES
1. t
1
= 15°c t
2
= 17°c
?t = t
2
– t
1
= 17 – 15 = 2°C = 2 + 273 = 275 K
m
v
= 100 g = 0.1 kg m
w
= 200 g = 0.2 kg
cu
g
= 420 J/kg–k W
g
= 4200 J/kg–k
(a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the 
vessel and water.
(b) Work done on the system = Heat produced unit
? dw = 100 × 10
–3
× 420 × 2 + 200 × 10
–3
× 4200 × 2 = 84 + 84 × 20 = 84 × 21 = 1764 J.
(c)dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]
2. (a) Heat is not given to the liquid. Instead the mechanical work done is converted 
to heat. So, heat given to liquid is z.
(b) Work done on the liquid is the PE lost by the 12 kg mass = mgh = 12 × 10× 
0.70 = 84 J
(c) Rise in temp at ?t We know, 84 = ms ?t 
? 84 = 1 × 4200 × ?t (for ‘m’ = 1kg) ? ?t = 
4200
84
= 0.02 k ?
V
P
f
T
P
A
B
T
P
A
B
V
P
12 kg
Laws of thermodynamics
26.3
3. mass of block = 100 kg
u = 2 m/s, m = 0.2 v = 0
dQ = du + dw 
In this case dQ = 0
? – du = dw ? du = ?
?
?
?
?
?
? ?
2 2
mu
2
1
mv
2
1
= 2 2 100
2
1
? ? ? = 200 J
4. Q = 100 J
We know, ?U = ?Q – ?W 
Here since the container is rigid, ?V = 0,
Hence the ?W = P ?V = 0,
So, ?U = ?Q = 100 J. ?
5. P
1
= 10 kpa = 10 × 10
3
pa. P
2
= 50 × 10
3
pa. v
1
= 200 cc. v
2
= 50 cc
(i) Work done on the gas = 
6 3
10 ) 200 50 ( 10 ) 50 10 (
2
1
?
? ? ? ? ? = – 4.5 J
(ii) dQ = 0 ? 0 = du + dw ? du = – dw = 4.5 J
6. initial State ‘I’ Final State ‘f’
Given 
1
1
T
P
= 
2
2
T
P
where P
1
? Initial Pressure ; P
2
? Final Pressure.
T
2
, T
1
? Absolute temp. So, ?V = 0
Work done by gas = P ?V = 0 ?
7. In path ACB, 
W
AC
+ W
BC
= 0 + pdv = 30 × 10
3
(25 – 10) × 10
–6
= 0.45 J
In path AB, W
AB
= ½ × (10 + 30) × 10
3
15 × 10
–6
= 0.30 J
In path ADB, W = W
AD
+ W
DB
= 10 × 10
3 
(25 – 10) × 10
–6
+ 0 = 0.15 J
8. ?Q = ?U + ?W 
In abc, ?Q = 80 J ?W = 30 J
So, ?U = (80 – 30) J = 50 J
Now in adc, ?W = 10 J
So, ?Q = 10 + 50 = 60 J [ ??U = 50 J]
9. In path ACB,
dQ = 50 0 50 × 4.2 = 210 J
dW = W
AC
+ W
CB
= 50 × 10
3
× 200 × 10
–6
= 10 J
dQ = dU + dW  
? dU = dQ – dW = 210 – 10 = 200 J
In path ADB, dQ = ?
dU = 200 J (Internal energy change between 2 points is always same)
dW = W
AD
+ W
DB
= 0+ 155 × 10
3
× 200 × 10
–6
= 31 J
dQ = dU + dW = 200 + 31 = 231 J = 55 cal
10. Heat absorbed = work done = Area under the graph
In the given case heat absorbed = area of the circle
= ? × 10
4
× 10
–6
× 10
3
= 3.14 × 10 = 31.4 J
D
P
V
C
10 kpa
B
25 cc
A
10 cc
30 kpa
D
V
P
C
200 cc
B
155 kpa
A
50 kpa
400 cc
d
V
P
c
b
a
P
V
(cc)
100
(kpa)
300
100
300
Page 4


26.1
CHAPTER 26
LAWS OF THERMODYNAMICS
QUESTIONS FOR SHORT ANSWER
1. No in isothermal process heat is added to a system. The temperature does not increase so the internal 
energy does not.
2. Yes, the internal energy must increase when temp. increases; as internal energy depends upon 
temperature U ? T
3. Work done on the gas is 0. as the P.E. of the container si increased and not of gas. Work done by the 
gas is 0. as the gas is not expanding.
The temperature of the gas is decreased.
4. W = F × d = Fd Cos 0° = Fd
Change in PE is zero. Change in KE is non Zero. 
So, there may be some internal energy.
5. The outer surface of the cylinder is rubbed vigorously by a polishing machine. 
The energy given to the cylinder is work. The heat is produced on the cylinder which transferred to the 
gas.
6. No. work done by rubbing the hands in converted to heat and the hands become warm.
7. When the bottle is shaken the liquid in it is also shaken. Thus work is done on the liquid. But heat is not 
transferred to the liquid.
8. Final volume = Initial volume. So, the process is isobaric.
Work done in an isobaric process is necessarily zero.
9. No word can be done by the system without changing its volume.
10. Internal energy = U = nC
V
T
Now, since gas is continuously pumped in. So n
2
= 2n
1
as the p
2
= 2p
1
. Hence the internal energy is 
also doubled.
11. When the tyre bursts, there is adiabatic expansion of the air because the pressure of the air inside is 
sufficiently higher than atmospheric pressure. In expansion air does some work against surroundings. 
So the internal energy decreases. This leads to a fall in temperature.
12. ‘No’, work is done on the system during this process. No, because the object expands during the 
process i.e. volume increases.
13. No, it is not a reversible process.
14. Total heat input = Total heat out put i.e., the total heat energy given to the system is converted to 
mechanical work.
15. Yes, the entropy of the body decreases. But in order to cool down a body we need another external sink 
which draws out the heat the entropy of object in partly transferred to the external sink. Thus once 
entropy is created. It is kept by universe. And it is never destroyed. This is according to the 2
nd
law of  
thermodynamics
OBJECTIVE – ?
1. (d) Dq = DU + DW. This is the statement of law of conservation of energy. The energy provided is 
utilized to do work as well as increase the molecular K.E. and P.E.
2. (b) Since it is an isothermal process. So temp. will remain constant as a result ‘U’ or internal energy will 
also remain constant. So the system has to do positive work.
3. (a) In case of A ?W
1
> ?W
2
(Area under the graph is higher for A than for B). 
?Q = ?u + dw.
du for both the processes is same (as it is a state function) 
??Q
1
> ?Q
2
  as ?W
1
> ?W
2
4. (b) As Internal energy is a state function and not a path function. ?U
1
= ?U
2
1
F
d 1
V
B ?Q 2 ?
P
A ?Q 1 ?
V
P
A
B
Laws of thermodynamics
26.2
5. (a) In the process the volume of the system increases continuously. Thus, the work 
done increases continuously.
6. (c) for A ? In a so thermal system temp remains same although heat is added. 
for B ? For the work done by the system volume increase as is consumes heat.
7. (c) In this case P and T varry proportionally i.e. P/T = constant. This is possible only 
when volume does not change. ? pdv = 0 ?
8. (c) Given : ?V
A
= ?V
B
. But P
A
< P
B
Now, W
A
= P
A
?V
B
; W
B
= P
B
?V
B
; So, W
A
< W
B
. 
?
9. (b) As the volume of the gas decreases, the temperature increases as well as the pressure. But, on 
passage of time, the heat develops radiates through the metallic cylinder thus T decreases as well as 
the pressure.
OBJECTIVE – ??
?
1. (b), (c) Pressure P and Volume V both increases. Thus work done is positive (V increases). Heat must 
be added to the system to follow this process. So temperature must increases.
2. (a) (b) Initial temp = Final Temp. Initial internal energy = Final internal energy.
i.e. ?U = 0, So, this is found in case of a cyclic process.
3. (d) ?U = Heat supplied, ?W = Work done.
( ?Q – ?W) = du, du is same for both the methods since it is a state function. ?
4. (a) (c) Since it is a cyclic process.
So, ?U
1
= – ?U
2
, hence ?U
1
+ ?U
2
= 0
?Q – ?W = 0 ?
5. (a) (d) Internal energy decreases by the same amount as work done. 
du = dw, ? dQ = 0. Thus the process is adiabatic. In adiabatic process, dU = – dw. Since ‘U’ decreases 
U
2
– U
2
is –ve. ?dw should be +ve ? ? ?
2 1
T T
1
nR
?
? ?
is +ve. T
1
> T
2
  ? Temperature decreases.
EXERCISES
1. t
1
= 15°c t
2
= 17°c
?t = t
2
– t
1
= 17 – 15 = 2°C = 2 + 273 = 275 K
m
v
= 100 g = 0.1 kg m
w
= 200 g = 0.2 kg
cu
g
= 420 J/kg–k W
g
= 4200 J/kg–k
(a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the 
vessel and water.
(b) Work done on the system = Heat produced unit
? dw = 100 × 10
–3
× 420 × 2 + 200 × 10
–3
× 4200 × 2 = 84 + 84 × 20 = 84 × 21 = 1764 J.
(c)dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]
2. (a) Heat is not given to the liquid. Instead the mechanical work done is converted 
to heat. So, heat given to liquid is z.
(b) Work done on the liquid is the PE lost by the 12 kg mass = mgh = 12 × 10× 
0.70 = 84 J
(c) Rise in temp at ?t We know, 84 = ms ?t 
? 84 = 1 × 4200 × ?t (for ‘m’ = 1kg) ? ?t = 
4200
84
= 0.02 k ?
V
P
f
T
P
A
B
T
P
A
B
V
P
12 kg
Laws of thermodynamics
26.3
3. mass of block = 100 kg
u = 2 m/s, m = 0.2 v = 0
dQ = du + dw 
In this case dQ = 0
? – du = dw ? du = ?
?
?
?
?
?
? ?
2 2
mu
2
1
mv
2
1
= 2 2 100
2
1
? ? ? = 200 J
4. Q = 100 J
We know, ?U = ?Q – ?W 
Here since the container is rigid, ?V = 0,
Hence the ?W = P ?V = 0,
So, ?U = ?Q = 100 J. ?
5. P
1
= 10 kpa = 10 × 10
3
pa. P
2
= 50 × 10
3
pa. v
1
= 200 cc. v
2
= 50 cc
(i) Work done on the gas = 
6 3
10 ) 200 50 ( 10 ) 50 10 (
2
1
?
? ? ? ? ? = – 4.5 J
(ii) dQ = 0 ? 0 = du + dw ? du = – dw = 4.5 J
6. initial State ‘I’ Final State ‘f’
Given 
1
1
T
P
= 
2
2
T
P
where P
1
? Initial Pressure ; P
2
? Final Pressure.
T
2
, T
1
? Absolute temp. So, ?V = 0
Work done by gas = P ?V = 0 ?
7. In path ACB, 
W
AC
+ W
BC
= 0 + pdv = 30 × 10
3
(25 – 10) × 10
–6
= 0.45 J
In path AB, W
AB
= ½ × (10 + 30) × 10
3
15 × 10
–6
= 0.30 J
In path ADB, W = W
AD
+ W
DB
= 10 × 10
3 
(25 – 10) × 10
–6
+ 0 = 0.15 J
8. ?Q = ?U + ?W 
In abc, ?Q = 80 J ?W = 30 J
So, ?U = (80 – 30) J = 50 J
Now in adc, ?W = 10 J
So, ?Q = 10 + 50 = 60 J [ ??U = 50 J]
9. In path ACB,
dQ = 50 0 50 × 4.2 = 210 J
dW = W
AC
+ W
CB
= 50 × 10
3
× 200 × 10
–6
= 10 J
dQ = dU + dW  
? dU = dQ – dW = 210 – 10 = 200 J
In path ADB, dQ = ?
dU = 200 J (Internal energy change between 2 points is always same)
dW = W
AD
+ W
DB
= 0+ 155 × 10
3
× 200 × 10
–6
= 31 J
dQ = dU + dW = 200 + 31 = 231 J = 55 cal
10. Heat absorbed = work done = Area under the graph
In the given case heat absorbed = area of the circle
= ? × 10
4
× 10
–6
× 10
3
= 3.14 × 10 = 31.4 J
D
P
V
C
10 kpa
B
25 cc
A
10 cc
30 kpa
D
V
P
C
200 cc
B
155 kpa
A
50 kpa
400 cc
d
V
P
c
b
a
P
V
(cc)
100
(kpa)
300
100
300
Laws of thermodynamics
26.4
11. dQ = 2.4 cal = 2.4 J Joules
dw = W
AB
+ W
BC
+ W
AC
= 0 + (1/2) × (100 + 200) × 10
3
200 × 10
–6
– 100 × 10
3
× 200 × 10
–6
= (1/2) × 300 × 10
3
200 × 10
–6
– 20 = 30 – 20 = 10 joules.
du = 0 (in a cyclic process)
dQ = dU +dW ? 2.4 J = 10 
? J = 
4 . 2
10
˜ 4.17 J/Cal.
12. Now, ?Q = (2625 × J) J
?U = 5000 J
From Graph ?W = 200 × 10
3
× 0.03 = 6000 J.
Now, ?Q = ?W + ?U 
? 2625 J = 6000 + 5000 J
J = 
2625
11000
= 4.19 J/Cal
?
13. dQ = 70 cal = (70 × 4.2) J
dW = (1/2) × (200 + 500) × 10
3
× 150 × 10
–6
= (1/2) × 500 × 150 × 10
–3
= 525 × 10
–1
= 52.5 J
dU = ? dQ = du + dw
? – 294 = du + 52.5 
? du = – 294 – 52.5 = – 346.5 J
14. U = 1.5 pV P = 1 × 10
5
Pa
dV = (200 – 100) cm
3
= 100 cm
3
= 10
–4
m
3
dU = 1.5 × 10
5
× 10
–4
  = 15
dW = 10
5
× 10
–4
= 10
dQ = dU + dW = 10 + 15 = 25 J
15. dQ = 10 J
dV = A × 10 cm
3
= 4 × 10 cm
3
= 40 × 10
–6
cm
3
dw = Pdv = 100 × 10
3
× 40 × 10
–6
= 4 cm
3
du = ? 10 = du + dw ? 10 = du + 4 ? du = 6 J.
16. (a) P
1
= 100 KPa 
V
1
  = 2 m
3
?V
1
= 0.5 m
3
?P
1
  = 100 KPa
From the graph, We find that area under AC is greater than area under 
than AB. So, we see that heat is extracted from the system.
(b) Amount of heat = Area under ABC. 
= 
5
10
10
5
2
1
? ? = 25000 J ?
17. n = 2 mole
dQ = – 1200 J
dU = 0 (During cyclic Process)
dQ = dU + dwc
? – 1200 = W
AB
+ W
BC
+ W
CA
? – 1200 = nR ?T + W
BC
+ 0
? – 1200 = 2 × 8.3 × 200 + W
BC
? W
BC
= – 400 × 8.3 – 1200 = – 4520 J. ?
P
V C
A
B
600 cc
200 kpa 100 kpa
700 cc
200 kpa
0.02 m
3
c
a
b
0.05 m
3
300 kpa
250 cc
100 cc
500 kpa 200 kpa
2 m
3
V
P
100 kpa
2.5 m
3
B
300 k
C
A
O
T
V
500 k
Page 5


26.1
CHAPTER 26
LAWS OF THERMODYNAMICS
QUESTIONS FOR SHORT ANSWER
1. No in isothermal process heat is added to a system. The temperature does not increase so the internal 
energy does not.
2. Yes, the internal energy must increase when temp. increases; as internal energy depends upon 
temperature U ? T
3. Work done on the gas is 0. as the P.E. of the container si increased and not of gas. Work done by the 
gas is 0. as the gas is not expanding.
The temperature of the gas is decreased.
4. W = F × d = Fd Cos 0° = Fd
Change in PE is zero. Change in KE is non Zero. 
So, there may be some internal energy.
5. The outer surface of the cylinder is rubbed vigorously by a polishing machine. 
The energy given to the cylinder is work. The heat is produced on the cylinder which transferred to the 
gas.
6. No. work done by rubbing the hands in converted to heat and the hands become warm.
7. When the bottle is shaken the liquid in it is also shaken. Thus work is done on the liquid. But heat is not 
transferred to the liquid.
8. Final volume = Initial volume. So, the process is isobaric.
Work done in an isobaric process is necessarily zero.
9. No word can be done by the system without changing its volume.
10. Internal energy = U = nC
V
T
Now, since gas is continuously pumped in. So n
2
= 2n
1
as the p
2
= 2p
1
. Hence the internal energy is 
also doubled.
11. When the tyre bursts, there is adiabatic expansion of the air because the pressure of the air inside is 
sufficiently higher than atmospheric pressure. In expansion air does some work against surroundings. 
So the internal energy decreases. This leads to a fall in temperature.
12. ‘No’, work is done on the system during this process. No, because the object expands during the 
process i.e. volume increases.
13. No, it is not a reversible process.
14. Total heat input = Total heat out put i.e., the total heat energy given to the system is converted to 
mechanical work.
15. Yes, the entropy of the body decreases. But in order to cool down a body we need another external sink 
which draws out the heat the entropy of object in partly transferred to the external sink. Thus once 
entropy is created. It is kept by universe. And it is never destroyed. This is according to the 2
nd
law of  
thermodynamics
OBJECTIVE – ?
1. (d) Dq = DU + DW. This is the statement of law of conservation of energy. The energy provided is 
utilized to do work as well as increase the molecular K.E. and P.E.
2. (b) Since it is an isothermal process. So temp. will remain constant as a result ‘U’ or internal energy will 
also remain constant. So the system has to do positive work.
3. (a) In case of A ?W
1
> ?W
2
(Area under the graph is higher for A than for B). 
?Q = ?u + dw.
du for both the processes is same (as it is a state function) 
??Q
1
> ?Q
2
  as ?W
1
> ?W
2
4. (b) As Internal energy is a state function and not a path function. ?U
1
= ?U
2
1
F
d 1
V
B ?Q 2 ?
P
A ?Q 1 ?
V
P
A
B
Laws of thermodynamics
26.2
5. (a) In the process the volume of the system increases continuously. Thus, the work 
done increases continuously.
6. (c) for A ? In a so thermal system temp remains same although heat is added. 
for B ? For the work done by the system volume increase as is consumes heat.
7. (c) In this case P and T varry proportionally i.e. P/T = constant. This is possible only 
when volume does not change. ? pdv = 0 ?
8. (c) Given : ?V
A
= ?V
B
. But P
A
< P
B
Now, W
A
= P
A
?V
B
; W
B
= P
B
?V
B
; So, W
A
< W
B
. 
?
9. (b) As the volume of the gas decreases, the temperature increases as well as the pressure. But, on 
passage of time, the heat develops radiates through the metallic cylinder thus T decreases as well as 
the pressure.
OBJECTIVE – ??
?
1. (b), (c) Pressure P and Volume V both increases. Thus work done is positive (V increases). Heat must 
be added to the system to follow this process. So temperature must increases.
2. (a) (b) Initial temp = Final Temp. Initial internal energy = Final internal energy.
i.e. ?U = 0, So, this is found in case of a cyclic process.
3. (d) ?U = Heat supplied, ?W = Work done.
( ?Q – ?W) = du, du is same for both the methods since it is a state function. ?
4. (a) (c) Since it is a cyclic process.
So, ?U
1
= – ?U
2
, hence ?U
1
+ ?U
2
= 0
?Q – ?W = 0 ?
5. (a) (d) Internal energy decreases by the same amount as work done. 
du = dw, ? dQ = 0. Thus the process is adiabatic. In adiabatic process, dU = – dw. Since ‘U’ decreases 
U
2
– U
2
is –ve. ?dw should be +ve ? ? ?
2 1
T T
1
nR
?
? ?
is +ve. T
1
> T
2
  ? Temperature decreases.
EXERCISES
1. t
1
= 15°c t
2
= 17°c
?t = t
2
– t
1
= 17 – 15 = 2°C = 2 + 273 = 275 K
m
v
= 100 g = 0.1 kg m
w
= 200 g = 0.2 kg
cu
g
= 420 J/kg–k W
g
= 4200 J/kg–k
(a) The heat transferred to the liquid vessel system is 0. The internal heat is shared in between the 
vessel and water.
(b) Work done on the system = Heat produced unit
? dw = 100 × 10
–3
× 420 × 2 + 200 × 10
–3
× 4200 × 2 = 84 + 84 × 20 = 84 × 21 = 1764 J.
(c)dQ = 0, dU = – dw = 1764. [since dw = –ve work done on the system]
2. (a) Heat is not given to the liquid. Instead the mechanical work done is converted 
to heat. So, heat given to liquid is z.
(b) Work done on the liquid is the PE lost by the 12 kg mass = mgh = 12 × 10× 
0.70 = 84 J
(c) Rise in temp at ?t We know, 84 = ms ?t 
? 84 = 1 × 4200 × ?t (for ‘m’ = 1kg) ? ?t = 
4200
84
= 0.02 k ?
V
P
f
T
P
A
B
T
P
A
B
V
P
12 kg
Laws of thermodynamics
26.3
3. mass of block = 100 kg
u = 2 m/s, m = 0.2 v = 0
dQ = du + dw 
In this case dQ = 0
? – du = dw ? du = ?
?
?
?
?
?
? ?
2 2
mu
2
1
mv
2
1
= 2 2 100
2
1
? ? ? = 200 J
4. Q = 100 J
We know, ?U = ?Q – ?W 
Here since the container is rigid, ?V = 0,
Hence the ?W = P ?V = 0,
So, ?U = ?Q = 100 J. ?
5. P
1
= 10 kpa = 10 × 10
3
pa. P
2
= 50 × 10
3
pa. v
1
= 200 cc. v
2
= 50 cc
(i) Work done on the gas = 
6 3
10 ) 200 50 ( 10 ) 50 10 (
2
1
?
? ? ? ? ? = – 4.5 J
(ii) dQ = 0 ? 0 = du + dw ? du = – dw = 4.5 J
6. initial State ‘I’ Final State ‘f’
Given 
1
1
T
P
= 
2
2
T
P
where P
1
? Initial Pressure ; P
2
? Final Pressure.
T
2
, T
1
? Absolute temp. So, ?V = 0
Work done by gas = P ?V = 0 ?
7. In path ACB, 
W
AC
+ W
BC
= 0 + pdv = 30 × 10
3
(25 – 10) × 10
–6
= 0.45 J
In path AB, W
AB
= ½ × (10 + 30) × 10
3
15 × 10
–6
= 0.30 J
In path ADB, W = W
AD
+ W
DB
= 10 × 10
3 
(25 – 10) × 10
–6
+ 0 = 0.15 J
8. ?Q = ?U + ?W 
In abc, ?Q = 80 J ?W = 30 J
So, ?U = (80 – 30) J = 50 J
Now in adc, ?W = 10 J
So, ?Q = 10 + 50 = 60 J [ ??U = 50 J]
9. In path ACB,
dQ = 50 0 50 × 4.2 = 210 J
dW = W
AC
+ W
CB
= 50 × 10
3
× 200 × 10
–6
= 10 J
dQ = dU + dW  
? dU = dQ – dW = 210 – 10 = 200 J
In path ADB, dQ = ?
dU = 200 J (Internal energy change between 2 points is always same)
dW = W
AD
+ W
DB
= 0+ 155 × 10
3
× 200 × 10
–6
= 31 J
dQ = dU + dW = 200 + 31 = 231 J = 55 cal
10. Heat absorbed = work done = Area under the graph
In the given case heat absorbed = area of the circle
= ? × 10
4
× 10
–6
× 10
3
= 3.14 × 10 = 31.4 J
D
P
V
C
10 kpa
B
25 cc
A
10 cc
30 kpa
D
V
P
C
200 cc
B
155 kpa
A
50 kpa
400 cc
d
V
P
c
b
a
P
V
(cc)
100
(kpa)
300
100
300
Laws of thermodynamics
26.4
11. dQ = 2.4 cal = 2.4 J Joules
dw = W
AB
+ W
BC
+ W
AC
= 0 + (1/2) × (100 + 200) × 10
3
200 × 10
–6
– 100 × 10
3
× 200 × 10
–6
= (1/2) × 300 × 10
3
200 × 10
–6
– 20 = 30 – 20 = 10 joules.
du = 0 (in a cyclic process)
dQ = dU +dW ? 2.4 J = 10 
? J = 
4 . 2
10
˜ 4.17 J/Cal.
12. Now, ?Q = (2625 × J) J
?U = 5000 J
From Graph ?W = 200 × 10
3
× 0.03 = 6000 J.
Now, ?Q = ?W + ?U 
? 2625 J = 6000 + 5000 J
J = 
2625
11000
= 4.19 J/Cal
?
13. dQ = 70 cal = (70 × 4.2) J
dW = (1/2) × (200 + 500) × 10
3
× 150 × 10
–6
= (1/2) × 500 × 150 × 10
–3
= 525 × 10
–1
= 52.5 J
dU = ? dQ = du + dw
? – 294 = du + 52.5 
? du = – 294 – 52.5 = – 346.5 J
14. U = 1.5 pV P = 1 × 10
5
Pa
dV = (200 – 100) cm
3
= 100 cm
3
= 10
–4
m
3
dU = 1.5 × 10
5
× 10
–4
  = 15
dW = 10
5
× 10
–4
= 10
dQ = dU + dW = 10 + 15 = 25 J
15. dQ = 10 J
dV = A × 10 cm
3
= 4 × 10 cm
3
= 40 × 10
–6
cm
3
dw = Pdv = 100 × 10
3
× 40 × 10
–6
= 4 cm
3
du = ? 10 = du + dw ? 10 = du + 4 ? du = 6 J.
16. (a) P
1
= 100 KPa 
V
1
  = 2 m
3
?V
1
= 0.5 m
3
?P
1
  = 100 KPa
From the graph, We find that area under AC is greater than area under 
than AB. So, we see that heat is extracted from the system.
(b) Amount of heat = Area under ABC. 
= 
5
10
10
5
2
1
? ? = 25000 J ?
17. n = 2 mole
dQ = – 1200 J
dU = 0 (During cyclic Process)
dQ = dU + dwc
? – 1200 = W
AB
+ W
BC
+ W
CA
? – 1200 = nR ?T + W
BC
+ 0
? – 1200 = 2 × 8.3 × 200 + W
BC
? W
BC
= – 400 × 8.3 – 1200 = – 4520 J. ?
P
V C
A
B
600 cc
200 kpa 100 kpa
700 cc
200 kpa
0.02 m
3
c
a
b
0.05 m
3
300 kpa
250 cc
100 cc
500 kpa 200 kpa
2 m
3
V
P
100 kpa
2.5 m
3
B
300 k
C
A
O
T
V
500 k
Laws of thermodynamics
26.5
18. Given n = 2 moles
dV = 0
in ad and bc.
Hence dW = dQ dW = dW
ab
+ dW
cd
= 
0
0
2
0
0
1
V 2
V
Ln nRT
V
V 2
Ln nRT ?
= nR × 2.303 × log 2(500 – 300)
= 2 × 8.314 × 2.303 × 0.301 × 200 = 2305.31 J
19. Given M = 2 kg 2t = 4°c Sw = 4200 J/Kg–k
?
0
= 999.9 kg/m
3
?
4
= 1000 kg/m
3
P = 10
5
Pa.
Net internal energy = dv
dQ = DU + dw ? ms ?Q ? = dU + P(v
0
– v
4
)
? 2 × 4200 × 4 = dU + 10
5
(m – m)
? 33600 = dU + 10
5
?
?
?
?
?
?
?
?
?
4 0
v
m
V
m
= dU + 10
5
(0.0020002 – 0.002) = dU + 10
5
0.0000002 
? 33600 = du + 0.02 ? du = (33600 – 0.02) J
20. Mass = 10g = 0.01kg.
P = 10
5
Pa
dQ = 
o H
2
Q 0° – 100° +  
o H
2
Q – steam
= 0.01 × 4200 × 100 + 0.01 × 2.5 × 10
6
= 4200 + 25000 = 29200
dW = P × ?V 
? = 
1000
01 . 0
6 . 0
01 . 0
? = 0.01699
dW = P ?V = 0.01699 × 10
5
1699J
dQ = dW + dU or dU = dQ – dW = 29200 – 1699 = 27501 = 2.75 × 10
4
J ?
21. (a) Since the wall can not be moved thus dU = 0 and dQ = 0.
Hence dW = 0.
(b) Let final pressure in LHS = P
1
In RHS = P
2
( ? no. of mole remains constant)
1
1
RT 2
V P
= 
RT 2
V P
1
? P
1
= 
1
1
T
T P
= 
?
?
2 1 2 1 1
T T ) P P ( P
As, T = 
?
?
2 1 2 1
T T ) P P (
Simillarly P
2
= 
?
? ) P P ( T P
2 1 1 2
(c) Let T
2
> T
1
and ‘T’ be the common temp.
Initially
2
V P
1
= n
1
rt
1
  ? n
1
= 
1
1
RT 2
V P
n
2
= 
2
2
RT 2
V P
Hence dQ = 0, dW = 0, Hence dU = 0.
In case (LHS) RHS
?u
1
= 1.5n
1
R(T - T
1
)  But ?u
1
- ?u
2
= 0 ?u
2
  = 1.5n
2
R(T
2
–T)
? 1.5 n
1
R(T -T
1
) = 1.5 n
2
R(T
2
–T) 
? n
2
T – n
1
T
1
= n
2
T
2
– n
2
T ? T(n
1
+ n
2
) = n
1
T
1
+ n
2
T
2
a
d
V
c
V
V 0 2V 0
500 k
200 k
b
V/2
U = 1.5nRT 
P 1 T 1 P 2 T 2
V/2